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Unformatted text preview: we only produce if S > 8 and we would shut
down if S < 8; hence S∗ = S ∗ = 8. If there is no cost of undoing options there will be no hysteresis
(e.g. producers will never be producing at a loss).
Question 17.14.
As in Example 17.1, we receive an underlying cash ﬂow worth 300/.03 = 10000 by paying a
strike of the mine cost plus the PV of the cost of extraction 250/.05 + 1000 = 6000. Note the
dividend implied by the lease rate is 10000 (.03) = 300 is equal to the interest cost on the strike
6000 (.05) = 300. This implies there is no value to waiting. Using the call option method,
CallPerpetual (10000, 6000, .0001, ln (1.05) , ln (1.03)) = {4000, 9903.68} . (12) The mine is worth its current NPV.
Question 17.16.
The value of a producing mine is
Vp (S) = S 250
S
− 5000 + PutPerpetual
,
, .2, ln (1.05) , ln (1.03)
.03
.03 .05 (13) which is the value of no shut down mine plus the value of being able to shutdown. Working
backwards, if S is the trigger price of our mine investment (i.e. we receive Vp S when St = S ),
the current value is
V S, S = 300
S h1 × Vp S − 1000 (14) Using h1 = 1.5812, we want to choose S to maximize the above quantity. Using Excel’s solver or
other numerical program we S = 464.00 and V 300, S = 4830.21. Note the trigger price is lower
than when there is no shut down option. 122 Chapter 18
The Lognormal Distribution
Question 18.2.
If z is standard normal, µ + σ × z is N µ, σ 2 hence our ﬁve standard normals can be use
to create the desired properties: .8 + 5 (−1.7) = −7.7, .8 + 5 (.55) = 3.55, .8 + 5 (−.3) = −0.7,
.8 + 5 (−.02) = .7, and .8 + 5 (.85) = 5.05.
Question 18.4.
Sums and differences of two random variables are normally distributed hence x1 + x2 is normally
distributed with mean µ1 + µ2 = 10 and variance
2
2
σ1 + σ2 + 2ρσ1 σ2 = .5 + 14 + 2 (.5) (14) (−.3) = 10.3 The difference is normally distributed with mean µ1 − µ2 = −6 and (higher) variance
2
2
σ1 + σ2 − 2ρσ1 σ2 = .5 + 14 + 2 (.5) (14) (.3) = 18.7. Question 18.6.
If x ˜N µ, σ 2 then E (ex ) = ex µ+.5σ ; using the given numbers, E (ex ) = e2+2.5 = 90.017. There
is a 50% probability x is below its mean of 2 hence the median of ex is e2 = 7.3891.
2 Question 18.8.
√
Since St = S0 exp α − 1 σ 2 t + σ tz where z is standard normal,
2
P (St > 105) = P
=P =P √
1
105
α − σ 2 t + σ tz > ln
2
100
z> ln −z < 105
100 − ln − α − 1σ2 t
2
√
σt
105
100 + α − 1σ2 t
2
√
σt = N (d2 ) √
1
where d2 = ln (100/105) + α − 2 σ 2 t / σ t . Using the given parameters, d2 = .045967and
N (d2 ) = .4817. For this parameter speciﬁcation, the probability St > 105 increases with t and
123 Part 5 Advanced Pricing Theory decreases with σ . Analytically, since N (d2 ) > 0, we have the derivative will have the same sign
as ∂d2 /∂t and ∂d2 /∂σ . Speciﬁcally,
α − σ 2 /2 − ln (S0 /K) /t
∂P (St > K)
= N (d2 )
>0
√
∂t
2σ t
since α − σ 2 /2 = .035 > 0 and ln (S0 /K) < 0. As an example, if t is 5 years, there is a 57.46%
chance of being greater than 105. For volatility, let t = 1. Then
∂P (St > K)
α − ln (S0 /K) 1
+
= −N (d2 )
2σ 2
4
∂σ <0 Question 18.10.
We have
P (St < 98) = P
=P √
1
98
α − σ 2 t + σ tz < ln
2
100
z< ln 98
100 − α − 1σ2 t
2
√
σt = N (−d2 ) with −d2 = (ln (98/100) − .035) /.3 = −.18401. Hence P (St < 98) = N (−.18401) = 42.70%.
Question 18.12.
See Figure Two. Option prices depend on the conditional (risk neutral) expectation, not the probability the option is in the money. As T increases, the likelihood that ST > KT may be lower; however,
the payoff depends on the conditional expectation (since the option does not pay a constant amount).
The increased dispersion offsets the lower probability (for the call option). 124 Chapter 18 The Lognormal Distribution
Figure Two (Problem 18.12)
0.9 0.8 0.7 P (ST < 100e.08T)
0.6 0.5 P (ST > 100e.08T)
0.4 0.3 0.2 0 5 10 T 15 20 25 Question 18.14.
The mean should be varying year by year; whereas, the standard deviation should be more stable. 125 Chapter 19
Monte Carlo Valuation
Question 19.2.
The histogram should be similar to a standard normal density (“bell” shaped). Since a uniform
2
distribution has a mean of 0.5 and a variance of 1/12, the mean of 1=1 ui − 6 is zero and the
i
variance (& standard deviation) will be one since
12 var ui = 12var (ui ) = 1. i =1 Question 19.4.
√
The standard deviation of the estimate will be sn / n where sn is the sample standard deviation of
the n simulations. Since sn is close to 2.9, n = 84000 should give a standard error close to 0.01.
Question 19.6.
The simulations should be generated by S1 = 100 exp .06 − .42 /2 + .4z where z is standard
normal. The claim prices should be e−.06 S α where α is the relevant power and the S α is the average
from the simulations. These values should be close to
100α exp (α − 1) .06 + α2
.4
2 . Using this, the three values (should be close to 12461...
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This document was uploaded on 03/11/2014 for the course FIN 402 at FPT University.
 Spring '14
 NguyenThiMaiLan
 Derivatives

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